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Discuss the differentiability of $e^{-|x|}$ ?

I tried something like making $2$ domains, $x\ge 0$ and $x<0$

$e^{-|x|} = e^{-x} | x\ge 0 $

and

$e^{-|x|} = e^{x} | x<0 $

By using definition of differentiability,

$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^+}\frac{e^{-x}-1}{x} = -1$$

and Similarly

$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^-}\frac{e^{x}-1}{x} =+1$$

Hence, I can say that it is not differentiable at $x=0$.

Is my understanding right or Am I missing something ?

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    $f(0)=1$ not $0$ – 2017-01-02
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    In both cases the lateral limits exists. Try to write $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$ and observe that $e^0\neq 0$. – 2017-01-02
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    Sorry, I missed that. I've corrected it now . – 2017-01-02

1 Answers 1

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Your guessing that $e^{-|x|}$ is not differentiable at $x=0$ is correct. Since $e^{0}=1$, we get $$ \lim_{x\to 0+}\frac{f(x)-f(0)}{x-0}=-1 $$ and $$ \lim_{x\to 0-}\frac{f(x)-f(0)}{x-0}=1. $$

  • When I first posted this, the original question was second revision and so I wrote that the reasoning has a flaw, but the original post was revised and the flaw has gone.